Page 140 - DMTH401_REAL ANALYSIS

Real Analysis

Notes when x  0.
0
Solution: The given function is not defined at x = 0 since f(0) = which is meaningless.
0
+
x 3
If x  0, then f(x) = . Therefore
2
Right Hand Limit = lim f(x)
+
x 0 0
-
+
(0 h) 3
+
= lim (h > 0)
-
h 0 2
= 3/2.
Left Hand Limit = lim f(x)
x 0 0
-
-
(0 h) 3
+
-
= lim f(x) = (h > 0)
-
h 0 2
= 3/2.
Since both the right hand and left hand limits exist and are equal,
lim f(x) = 3/2.
x 0
-
We, now, discuss the theorem concerning the existence of limit and that of right and the left hand
limits.
Theorem 2: Let f be a real function. Then
lim f(x) = A if and only if lim f(x) and lim f(x)
-
-
-
x a x a + x a –
both exist and are equal to A.
Proof: If lim f(x) = A, then corresponding to any  > 0, there exists a  > 0 such that
-
x a +
|f(x) – A| <  whenever 0 < |x – a| < 
i.e., | f(x) – A| <  whenever a –  < x < a + , x  a
This implies that |f(x) – A| <  whenever a –  < x < a
and |f(x) – A| <  whenever a < x < a + .
Hence both the left hand and right hand limits exist and are equal to A. Conversely, if f(a+) and
f(a–) exist and are equal to A say, then corresponding to E > 0, there exist positive numbers  and
1
 such that
2
|f(x) – A | < E whenever a < x < a + 
1
and
|f(x) – A| <  whenever a –  < x < a.
2
Let 6 be the minimum of  and  . Then
1 2
|f(x) – A| < E whenever a –  < x < a + , x  a
i.e. |f(x) – A| <  if 0 < |x – a| < 
which proves that

lim f(x) exists and lim f(x) = A.
-
-
x a x a

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